2018 AIME II Problems/Problem 14

Problem

The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ , $PB = 4$ , $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Diagram

$[asy] size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); pair A,B,C,P,Q,Wp,X,Y,Z; B=origin; C=(6.75,0); A=IP(CR(B,7),CR(C,8)); path c=incircle(A,B,C); Wp=IP(c,A--C); Z=IP(c,A--B); X=IP(c,B--C); Y=IP(c,A--X); pair I=incenter(A,B,C); P=extension(A,B,Y,Y+dir(90)*(Y-I)); Q=extension(A,C,P,Y); draw(A--B--C--cycle, black+1); draw(c^^A--X^^P--Q); pen p=4+black; dot("$A$",A,N,p); dot("$B$",B,SW,p); dot("$C$",C,SE,p); dot("$X$",X,S,p); dot("$Y$",Y,dir(55),p); dot("$W$",Wp,E,p); dot("$Z$",Z,W,p); dot("$P$",P,W,p); dot("$Q$",Q,E,p); MA("\beta",C,X,A,0.3,black); MA("\alpha",B,A,X,0.7,black); [/asy]$

Solution 1

Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$ , respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$ . Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$ , it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$ . By equal tangents, $PZ = PY$ . Applying the Law of Sines to $\triangle APY$ yields $\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.$ Similarly, applying the Law of Sines to $\triangle ABX$ gives $\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.$ It follows that $2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,$ implying $AZ = \tfrac{21}5$ . Applying the same argument to $\triangle AQY$ yields $2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),$ from which $AQ = \tfrac{168}{59}$ . The requested sum is $168 + 59 = \boxed{227}$ .

Solution 2 (Projective)

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $Z$ and $W$ . By Brianchon's theorem on tangential hexagons $QWCBZP$ and $PYQCXB$ , we know that $ZW,CP,BQ$ and $XY$ are concurrent at a point $O$ . Let $PQ \cap BC = M$ . Then by La Hire's $A$ lies on the polar of $M$ so $M$ lies on the polar of $A$ . Therefore, $ZW$ also passes through $M$ . Then projecting through $M$ , we have $-1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).$ Therefore, $\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1$ . Since $ZB+ZP=4$ we know that $ZP = \frac{6}{5}$ and $ZB = \frac{14}{5}$ . Therefore, $AW = AZ = \frac{21}{5}$ and $WC = 8 - \frac{21}{5} = \frac{19}{5}$ . Since $(A,W;Q,C) = -1$ , we also have $\frac{AQ \cdot WC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$ . Solving for $AQ$ , we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$ . 😃 -Vfire

Solution 3 (Combination of Law of Sine and Law of Cosine)

Let the center of the incircle of $\triangle ABC$ be $O$ . Link $OY$ and $OX$ . Then we have $\angle OYP=\angle OXB=90^{\circ}$

$\because$ $OY=OX$

$\therefore$ $\angle OYX=\angle OXY$

$\therefore$ $\angle PYX=\angle YXB$

$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$ , let $MP=YP=x$ and $NQ=YQ=y$ .

Use Law of Sine in $\triangle APY$ and $\triangle AXB$ , we have

$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$

$\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$

therefore we have

$\frac{3}{x}=\frac{7}{4-x}$

Solve this equation, we have $x=\frac{6}{5}$

As a result, $MB=4-x=\frac{14}{5}=BX$ , $AM=x+3=\frac{21}{5}=AN$ , $NC=8-AN=\frac{19}{5}=XC$ , $AQ=\frac{21}{5}-y$ , $PQ=\frac{6}{5}+y$

So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$

Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$ , we have

$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$

$\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

And we have

$\cos \angle BAC=\cos \angle PAQ$

So

$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

Solve this equation, we have $y=\frac{399}{295}=QN$

As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$

So, the final answer of this question is $168+59=\boxed {227}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 4 (Projective geometry)

Claim

Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $M$ and $N$ , respectively. Then lines $PQ, MN,$ and $BC$ are concurrent and lines $PC, MN, AX,$ and $BQ$ are concurrent.

Proof

Let $E$ be point of crossing $AX$ and $BQ.$ We make projective transformation such that circle $\omega$ maps into the circle and point $E$ maps into the center of new circle point $I.$ We denote images using notification $X \rightarrow X'.$

$BCQP$ maps into $B'C'Q'P'$ , so lines $B'Q'$ and $A'X'$ be the diameters. This implies $P'Q'||B'C', \angle B'P'Q' = \angle B'C'Q' = 90^\circ \implies B'C'Q'P'$ be a square.

Therefore $M'N'$ be the diameter $\implies P'C', B'Q',$ be diagonals of the square. $M'N'$ and $X'Y'$ be midlines which crossing in the center $I.$ Therefore lines $PC, MN, AX,$ and $BQ$ are concurrent.

Lines $P'Q'||M'N' ||B'C' \implies PQ, MN$ and $BC$ are concurrent.

Solution The cross-ratio associated with a list of four collinear points $A,P,M,D$ is defined as $(A,P;M,B)={\frac {AP\cdot MB}{AB\cdot PM}}.$ The cross-ratio be projective invariant of a quadruple of collinear points, so

$(A,P; M,B) = {\frac {A'P'\cdot M'B'}{A'B'\cdot P'M'}} = \frac {M'B'}{P'M'} = 1.$ $(A,P; M,B)={\frac {3\cdot (7 - AM)}{7\cdot (AM - 3)}} = 1 \implies AM = \frac {21}{5} \implies AN = AM = \frac {21}{5}.$

$(A,Q;N,C)={\frac {AQ\cdot NC}{AC\cdot QN}} = \frac {AQ\cdot (AC- AN)}{AC\cdot (AN-AQ)} = 1.$ $AQ \cdot (8 - \frac{21}{5}) = 8 \cdot (\frac{21}{5} – AQ) \implies AQ = \frac{168}{59}.$

For visuals only, I will show how one can find the perceptor $D$ and the image’s plane. $E_0$ is image of inversion $E$ with respect $\omega.$ $UW$ is the diameter of $\omega, E,E_0,U,W$ are collinear. $DU \perp \omega, DE_0\perp WD, UV \perp WD, UV$ is diameter of $\omega'$ .

Plane of images is perpendicular to $WD.$

Last diagram shows the result of transformation. Transformation is possible. The end.

vladimir.shelomovskii@gmail.com, vvsss

Solution 5

Firstly, assume $PY=x=PZ, ZB=4-x=BX, AZ=AW=3+x, CW=CX=5-x, QY=QW=y, AQ=3+x-y$

By tangency, we have $\angle{PYZ}=\angle{PZY}=\angle{YXZ}; \angle{BZX}=\angle{BXZ}=\angle{ZYX}; \angle{PYX}=\angle{BXY}$

Similar reason yields $\angle{QYX}=\angle{CXY}$ . Apply Law of sines

We have $\frac{3}{\sin{\angle{PYA}}}=\frac{x}{\sin{\angle{PAY}}}, \frac{7}{\sin{\angle{AXB}}}=\frac{4-x}{\sin{\angle{BAX}}}$ Since $\angle{PYA}=180-\angle{BXA}$ so their sine values would be the same. Solve this system and we have $\frac{3}{7}=\frac{x}{4-x}, x=\frac{6}{5}$

Apply the same process in $\triangle{AQY}, \triangle{AXC}$ , we have $\frac{3+x-y}{8}=\frac{y}{5-x}, y=\frac{399}{295}$

The desired length is $3+x-y=\frac{168}{59}\implies \boxed{227}$

~Bluesoul

Video Solution by Mop 2024

https://youtu.be/SIs1JFLFzyw

~r00tsOfUnity

2018 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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